How do you balance AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3?

Jan 18, 2016

$2 A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \rightarrow 6 K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

Explanation:

To balance this equation, all you have to do is make sure that there is an equal number of each element in both the reactant and product side.

Step by step balancing:

$A l B {r}_{3} + {K}_{2} S {O}_{4} \to K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ (unbalanced)

reactants:
Al - 1 Br - 3 K - 2 S - 1 O - 4
products:
Al - 2 Br - 1 K - 1 S - 3 O - 12

$A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ (unbalanced)

reactants:
Al - 1 Br - 3 K - 6 S - 3 O - 12
products:
Al - 2 Br - 1 K - 1 S - 3 O - 12

$2 A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ (unbalanced)

reactants:
Al - 2 Br - 6 K - 6 S - 3 O - 12
products:
Al - 2 Br - 1 K - 1 S - 3 O - 12

$2 A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to 6 K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ (balanced)

reactants:
Al - 2 Br - 6 K - 6 S - 3 O - 12
products:
Al - 2 Br - 6 K - 6 S - 3 O - 12